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(B) According to Rayleigh's criterion, the angular separation between the central maximum and the first minimum of a diffraction pattern for a circula

Vedclass · Accepted Answer

$3.05 	imes 10^{-4} 	ext{ cm}$

Vedclass · Answer

(B) According to Rayleigh's criterion, the angular separation between the central maximum and the first minimum of a diffraction pattern for a circular aperture is given by $	heta = 1.22 \lambda / D$.Here, $\lambda = 5000 	ext{ Å} = 5000 	imes 10^{-8} 	ext{ cm} = 5 	imes 10^{-5} 	ext{ cm}$.The diameter of the lens is $D = 5 	ext{ cm}$.The angular separation is $	heta = 1.22 	imes (5 	imes 10^{-5} 	ext{ cm}) / 5 	ext{ cm} = 1.22 	imes 10^{-5} 	ext{ radians}$.The linear distance $x$ in the focal plane is given by $x = f 	heta$, where $f = 25 	ext{ cm}$.$x = 25 	ext{ cm} 	imes 1.22 	imes 10^{-5} 	ext{ rad} = 30.5 	imes 10^{-5} 	ext{ cm} = 3.05 	imes 10^{-4} 	ext{ cm}$.

$A$ converging lens of a telescope with a diameter of $5 \text{ cm}$ has a focal length of $25 \text{ cm}$. In the focal plane of the lens, the distance between the centres of the Fraunhofer diffraction pattern is (wavelength of light used = $5000 \text{ Å}$, Rayleigh's criterion is satisfied).

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